Two masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the center of mass of the system will be ( $g =$ acceleration due to gravity)

\frac{g (m_{1} - m_{2})}{(m_{1} + m_{2})}

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\frac{g (m_{1} - m_{2})^{2}}{(m_{1} + m_{2})^{2}}

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\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

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\frac{g (m_{1} + m_{2})^{2}}{(m_{1} - m_{2})}

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Solution

The correct option is A $\frac{g (m_{1} - m_{2})^{2}}{(m_{1} + m_{2})^{2}}$
$\begin{matrix} a_{c o m} = \frac{m_{1} a - m_{2} a}{m_{1} + m_{2}} = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}} . \frac{(m_{1} - m_{2}) g}{(m_{1} + m_{2})} = g \frac{{(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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Two masses m1 and m2 (m1>m2) are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the center of mass of the system will be (g=acceleration due to gravity)

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Two masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the center of mass of the system will be ( $g =$ acceleration due to gravity)